Is Ampere's law consistent with the general rule (Eq. 1.46) that divergence-of-curl is always zero? Show that Ampere's law cannot be valid, in general, outside magnetostatics. Is there any such "defect" in the other three Maxwell equations?

The divergence of curl is always zero.

$\nabla (\nabla .v)=0$

Write all the Maxwell equations:

$\begin{array}{c}\mathbf{\nabla }\mathbf{\times }\mathbf{E}\mathbf{=}\frac{\mathbf{\rho }}{{\mathbf{\epsilon }}_{\mathbf{0}}}\\ \mathbf{\nabla }\mathbf{\times }\mathbf{B}\mathbf{=}\mathbf{0}\\ \mathbf{\nabla }\mathbf{\times }\mathbf{E}\mathbf{=}\frac{\mathbf{-}\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{t}}\\ \mathbf{\nabla }\mathbf{\times }\mathbf{B}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{J}\end{array}$

Here,role="math" localid="1658226949122" $\mathit{E}$ is the electric field, $\mathit{B}$is the magnetic field, $\mathit{\rho }$is the charge density, ${\mathit{\epsilon }}_{\mathbf{0}}$permittivity of space, ${\mathit{\mu }}_{\mathbf{0}}$is the permeability of the space and $\mathit{J}$is the current density.

The expression for the continuity equation is given by,

$\begin{array}{l}\nabla \cdot J+\frac{d\rho }{dt}=0\\ \nabla \cdot J=-\frac{d\rho }{dt}\end{array}$

The expression for the ampere’s law is given by,

$\nabla \times B={\mu }_{0}J$

Take the divergence of both the sides of the above equation.

$\begin{array}{l}\nabla \cdot (\nabla \times B)=\nabla \cdot \left({\mu }_{0}J\right)\\ \nabla \cdot (\nabla \times B)={\mu }_0(\nabla \cdot J)\end{array}$

Substitute$-\frac{d\rho }{dt}$for $\nabla \cdot J$into above equation.

From the above it is proved that the value of curl is not equal to zero unless the charge density is constant quantity.

Therefore, it is proved that the ampere’s law is not valid for outside the magnetostatics.

Consider the Maxwell’s equations as,

$\nabla \times E=-\frac{\partial B}{\partial t}$

Take the divergence of both the sides of the above equation.

Substitute for into above equation.

$\begin{array}{l}\nabla \cdot (\nabla \times E)=\nabla \cdot (-\frac{\partial B}{\partial t})\\ \nabla \cdot (\nabla \times E)=\nabla \cdot \frac{d}{dt}B\\ \nabla \cdot (\nabla \times E)=\frac{d}{dt}(\nabla \cdot B)\end{array}$

Substitute$0$for$\nabla \cdot B$into above equation.

$\begin{array}{l}\nabla \cdot (\nabla \times E)=\frac{d}{dt}\left(0\right)\\ \nabla \cdot (\nabla \times E)=0\end{array}$

Hence from the above, consistence of curl is zero.

The two equations are $\nabla \cdot E=\frac{\rho }{{\epsilon }_o}$and $\nabla \cdot B=0$are the divergence equation and can’t be vanished by any form.

Therefore, defect in this equation is not consistence with the general rule of Eq. 1.46.

Hence the divergence of curl is not always equal to zero and it is proved that the ampere’s law is not valid outside the magnetostatics. The defect in this equation is not consistence in general rule.